# Check if a number can be represented as sum of two consecutive perfect cubes

Given an integer N, the task is to check if this number can be represented as the sum of two consecutive perfect cubes or not.

Examples:

Input: N = 35
Output: Yes
Explanation:
Since, 35 = 23 + 33, therefore the required answer is Yes.

Input: N = 14
Output: No

Naive Approach: The simplest approach to solve the problem is to iterate from 1 to cube root of N and check if the sum of perfect cubes of any two consecutive numbers is equal to N or not. If found to be true, print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

## C++

 // C++ Program of the // above approach   #include  using namespace std;   // Function to check if a number // can be expressed as the sum of // cubes of two consecutive numbers bool isCubeSum(int n) {     for (int i = 1; i * i * i <= n; i++) {         if (i * i * i                 + (i + 1) * (i + 1) * (i + 1)             == n)             return true;     }     return false; }   // Driver Code int main() {     int n = 35;       if (isCubeSum(n))         cout << "Yes";     else         cout << "No"; }

## Java

 // Java program of the // above approach import java.util.*;   class GFG{   // Function to check if a number // can be expressed as the sum of // cubes of two consecutive numbers static boolean isCubeSum(int n) {     for(int i = 1; i * i * i <= n; i++)     {         if (i * i * i + (i + 1) *               (i + 1) * (i + 1) == n)             return true;     }     return false; }   // Driver Code public static void main(String[] args) {     int n = 35;       if (isCubeSum(n))         System.out.print("Yes");     else         System.out.print("No"); } }   // This code is contributed by Amit Katiyar

## Python3

 # Python3 program of the # above approach   # Function to check if a number # can be expressed as the sum of # cubes of two consecutive numbers def isCubeSum(n):           for i in range(1, int(pow(n, 1 / 3)) + 1):         if (i * i * i + (i + 1) *               (i + 1) * (i + 1) == n):             return True;       return False;   # Driver Code if __name__ == '__main__':           n = 35;       if (isCubeSum(n)):         print("Yes");     else:         print("No");   # This code is contributed by Amit Katiyar

## C#

 // C# program of the // above approach using System;   class GFG{   // Function to check if a number // can be expressed as the sum of // cubes of two consecutive numbers static bool isCubeSum(int n) {     for(int i = 1; i * i * i <= n; i++)     {         if (i * i * i + (i + 1) *               (i + 1) * (i + 1) == n)             return true;     }     return false; }   // Driver Code public static void Main(String[] args) {     int n = 35;       if (isCubeSum(n))         Console.Write("Yes");     else         Console.Write("No"); } }   // This code is contributed by Amit Katiyar

## Javascript

 

Output:

Yes

Time Complexity: O(N1/3)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following observations:

• A number can be represented as the sum of the perfect cube of two consecutive numbers if the sum of the cube root of both consecutive numbers is equal to N.
• This can be checked by the formula: • For example, if N = 35, then check if the equation below is equal to N or not: Below is the implementation of the above approach:

## C++

 // C++ Program to // implement above approach   #include  using namespace std;   // Function to check that a number // is the sum of cubes of 2 // consecutive numbers or not bool isSumCube(int N) {     int a = cbrt(N);     int b = a - 1;       // Condition to check if a     // number is the sum of cubes of 2     // consecutive numbers or not     return ((a * a * a + b * b * b) == N); }   // Driver Code int main() {     int i = 35;     // Function call     if (isSumCube(i)) {         cout << "Yes";     }     else {         cout << "No";     }     return 0; }

## Java

 // Java program to implement // above approach class GFG{   // Function to check that a number // is the sum of cubes of 2 // consecutive numbers or not static boolean isSumCube(int N) {     int a = (int)Math.cbrt(N);     int b = a - 1;       // Condition to check if a     // number is the sum of cubes of 2     // consecutive numbers or not     return ((a * a * a + b * b * b) == N); }   // Driver Code public static void main(String[] args) {     int i = 35;           // Function call     if (isSumCube(i))      {         System.out.print("Yes");     }     else     {         System.out.print("No");     } } }   // This code is contributed by Amit Katiyar

## Python3

 # Python3 program to  # implement above approach    # Function to check that a number  # is the sum of cubes of 2 # consecutive numbers or not def isSumCube(N):       a = int(pow(N, 1 / 3))     b = a - 1       # Condition to check if a     # number is the sum of cubes of 2     # consecutive numbers or not     ans = ((a * a * a + b * b * b) == N)       return ans   # Driver Code i = 35   # Function call if(isSumCube(i)):     print("Yes") else:     print("No")   # This code is contributed by Shivam Singh

## C#

 // C# program to implement // above approach using System; class GFG{   // Function to check that a number // is the sum of cubes of 2 // consecutive numbers or not static bool isSumCube(int N) {   int a = (int)Math.Pow(N, (double) 1 / 3);   int b = a - 1;     // Condition to check if a   // number is the sum of cubes of 2   // consecutive numbers or not   return ((a * a * a + b * b * b) == N); }   // Driver Code public static void Main(String[] args) {   int i = 35;       // Function call   if (isSumCube(i))    {     Console.Write("Yes");   }   else   {     Console.Write("No");   } } }   // This code is contributed by 29AjayKumar

## Javascript

 

Output:

Yes

Time Complexity: O(logN)  because using cbrt function
Auxiliary Space: O(1)

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